Question

A 6.00 L tank at 10.6°C contains 12.2 g of N2O(g) and 8.28 g of CO2(g), what are the partial pressures of N2O and CO2?

Answer 1

Molar mass of N2O = 2*MM(N) + 1*MM(O)

= 2*14.01 + 1*16.0

= 44.02 g/mol

Molar mass of CO2 = 1*MM(C) + 2*MM(O)

= 1*12.01 + 2*16.0

= 44.01 g/mol

n(N2O) = mass of N2O/molar mass of N2O

= 12.2/44.02

= 0.2771

n(CO2) = mass of CO2/molar mass of CO2

= 8.28/44.01

= 0.1881

n(N2O),n1 = 0.2771 mol

n(CO2),n2 = 0.1881 mol

Total number of mol = n1+n2

= 0.2771 + 0.1881

= 0.4653 mol

we have:

V = 6.0 L

n = 0.4653 mol

T = 10.6 oC

= (10.6+273) K

= 283.6 K

we have below equation to be used:

P * V = n*R*T

P * 6 L = 0.4653 mol* 0.0821 atm.L/mol.K * 283.6 K

P = 1.806 atm

Partial pressure of each components are

p(N2O),p1 = (n1*Ptotal)/total mol

= (0.2771 * 1.806)/0.4653

= 1.08 atm

p(CO2),p2 = (n2*Ptotal)/total mol

= (0.1881 * 1.806)/0.4653

= 0.7303 atm

partial pressure of N2O = 1.08 atm

partial pressure of CO2 = 0.730 atm